# Question #b2d74

##### 1 Answer

#### Explanation:

I'll show you how to solve this problem **without** using the formula given to you first, then double-check the result by using the formula.

Start by writing the balanced chemical equation that describes this combustion reaction

#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> color(blue)(3)"CO"_ (2(g)) + 4"H"_ 2"O"_ ((l))#

Notice that when **mole** of propane reacts, **moles** of carbon dioxide are produced. To convert this to *grams*, use the **molar masses** of propane and carbon dioxide

#1 color(red)(cancel(color(black)("mole C"_3"H"_8))) * "44.10 g"/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = "44.10 g"#

#3 color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "132.03 g"#

So, this tells you that **for every** **theoretically** produces

I say *theoretically* because this corresponds to a

Now, you are told that the reaction has a **for every** *theoretically produced*, the reaction *actually produces*

Use this to find the mass of carbon dioxide that would correspond to a

#8.53 color(red)(cancel(color(black)("g CO"_2))) * ("100 g CO"_2color(white)(a)"theoretical")/(93color(red)(cancel(color(black)("g CO"_2color(white)(a)"actual")))) = "9.172 g CO"_2#

Now all you have to do is use the **gram ratio** that exists between propane and carbon dioxide to see what mass of propane would be needed to produce this much carbon dioxide

#9.172 color(red)(cancel(color(black)("g CO"_2))) * ("44.10 g C"_3"H"_8)/(132.03color(red)(cancel(color(black)("g CO"_2)))) = color(green)(bar(ul(|color(white)(a/a)color(black)("3.06 g C"_3"H"_8)color(white)(a/a)|)))#

I'll leave the answer rounded to three **sig figs**.

So, you have this formula

#color(purple)(bar(ul(|color(white)(a/a)color(black)("% yield" = "actual yield"/"theoretical yield" xx 100%)color(white)(a/a)|)))#

You know the percent yield and the actual yield, so your job here is to figure out what the *theoretical yield* would be.

Rearrange to find

#"theoretical yield" = "actual yield"/"% yield" xx 100%#

Plug in your values to find

#"theoretical yield" = "8.53 g"/(93color(red)(cancel(color(black)(%)))) xx 100color(red)(cancel(color(black)(%))) = "9.172 g"#

Once again, you know that when this reaction has a **theoretically produce**

At this point, you would use the gram ratio again to see how much propane would be needed to produce

As a final note, the general idea here is that the amount of propane needed to produce